Exercise sheet 8
not to be handed in
solutions
dy
− y = e2x .
1. Solve dx
Using integrating factors. Multiply by M to obtain
M
dy
− M y = M e2x .
dx
(1)
We want the equation to be equal to
(M y)0 = M e2x ,
(2)
which by the product rule is
M
dy
dM
+
y = M e2x .
dx
dx
(3)
Comparing (1) and (3) we then see that we must choose the integrating
factor M to satisfy
dM
= −M.
dx
To solve this we could use that it is a separable equation, but here we
easily see a solution to be M = e−x . So our equation for y becomes
e−x
dy
− e−x y = ex ,
dx
when we write it in the form of (3), or more appropriately it becomes
(e−x y)0 = ex
when we write it in the form of (2). We simply integrate this last equation
to obtain
e−x y = ex + C
(remember to put the constant in at this point) and solve for y to obtain
y = e2x + Cex .
Using variation of parameters. We first solve the equation
dz
− z = 0.
dx
This can be done by separation of variables, but in this case it is easy
enough to see that z = ex is a solution. We then substitute y = Cex
into our equation to obtain an equation for C. Using the product rule we
obtain
C 0 ex + Cex − Cex = e2x .
Two terms cancel each other (as always happens when using this method)
so that we obtain
C 0 ex = e2x .
We divide by ex to obtain
C 0 = ex
and integrate to obtain
C = ex + K.
Since y = Cex we then have y = e2x + Kex .
Using the method of undetermined coefficients. We first solve the equation
dz
dx − z = 0. This can be done by separation of variables, but in this case
it is easy enough to see that z = Cex is the general solution. We then
try a solution of the same form as the right-hand side, in this case we try
y = Ae2x . Substituting this into the equation gives
2Ae2x − Ae2x = e2x .
It follows that A = 1, so that y = e2x is a solution and y = e2x + Cex is
the general solution.
dy
2. Solve dx
+ y tan x = sin 2x with initial condition y(0) = 1.
Using integrating factors. Multiply by M to obtain
M
dy
+ yM tan x = M sin 2x.
dx
(4)
We want the equation to be equal to
(M y)0 = M sin 2x,
(5)
which by the product rule is
M
dy
dM
+
y = M sin 2x.
dx
dx
(6)
Comparing (4) and (6) we then see that we must choose the integrating
factor M to satisfy
dM
= M tan x.
dx
To solve this we use that this is a separable equation. We separate variables to obtain
Z
Z
dM
= tan x dx.
M
2
We now carry out the integrations to obtain (we can omit the constant
since we only need one integrating factor):
ln |M | = − ln | cos x|,
(7)
sin x
where we use that tan x = cos
x and used substitution with u = cos x so
du
that dx = − sin x to obtain
Z
Z
−1
tan x dx =
du = − ln |u| + C = − ln | cos x| + C,
u
and we subsequently take C = 0 to obtain the simplest integration factor.
Using the logarithmic rules we write (7) as
1
,
ln |M | = ln
| cos x|
so that we see that M =
y becomes
1
cos x
is an integrating factor. So our equation for
1 dy
tan x
sin 2x
−y
=
,
cos x dx
cos x
cos x
when we write it in the form of (3), or more appropriately it becomes
(
1
sin 2x
y)0 =
.
cos x
cos x
when we write it in the form of (2). We use the double angle formula
for the sine to write the right-hand side as 2 sin x, so that we obtain the
equation
1
(
y)0 = 2 sin x,
cos x
which we simply integrate to obtain
1
y = −2 cos x + C.
cos x
(remember to put the constant in at this point) and solve for y to obtain
y = −2 cos2 x + C cos x.
We now determine the arbitrary constant from the initial value. We have
y(0) = −2 + C = 1,
so that C = 3. The unique solution then is y = −2 cos2 x + 3 cos x.
Using variation of parameters. We first solve the equation
dz
+ z tan x = 0.
dx
3
This can be done by separation of variables as follows.
Z
Z
dz
= − tan x dx.
z
We obtain from this
ln |z| = ln | cos x|,
(8)
sin x
where we use that tan x = cos
x and used substitution with u = cos x so
du
that dx = − sin x to obtain
Z
Z
1
− tan x dx =
du = ln |u| + K = ln | cos x| + K,
u
and we subsequently take K = 0 to obtain the simplest solution. From
(8) it follows that z = cos x. We then substitute y = C cos x into our
equation to obtain an equation for C. Using the product rule we obtain
C 0 cos x − C sin x + C cos x tan x = sin 2x.
Two terms cancel each other (as always happens when using this method)
so that we obtain
C 0 cos x = sin 2x.
We divide by cos x and use the double angle formula for the sine to obtain
C 0 = 2 sin x
and integrate to obtain
C = −2 cos x + K.
Since y = C cos x we then have y = −2 cos2 x+K cos x. We now determine
the arbitrary constant from the initial value. We have
y(0) = −2 + K = 1,
so that K = 3. The unique solution then is y = −2 cos2 x + 3 cos x.
The method of undetermined coefficients is not really applicable in this
case since tan x appears as the coefficient of y in the differential equation
(this coefficient really needs to be constant for the method of undetermined
coefficients to work properly).
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